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  • If the sum of an infinite decreasing G.P is 3 and sum of the cubes of its terms is 108/13 , then common ratio is given by

If the sum of an infinite decreasing G.P is 3 and sum of the cubes of its terms is 108/13 , then common ratio is given by

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Solution:     

             a/1-r=3   ,   a^3/(1-r^3)=108/13  Where left | r ight |< 1

           herefore     27(1-r)^3/1-r^3=108/13

       Rightarrow 1-2r+r^2/1+r+r^2=4/13

     Rightarrow     13-26r+13r^2=4+4r+4r^2

     Rightarrow    9r^2-30r+9=0Rightarrow 3r^2-10r+3=0

    Rightarrow      (3r-1)(r-3)=0Rightarrow r=1/3

Posted by

Deependra Verma

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